(a) Temperature of interface
$\theta = \frac{K_{1}\theta_{1}l_{2} + K_{2}\theta_{2}l_{1}}{K_{1}l_{2} + K_{2}l_{1}}$
It is given that $K_{Cu} = 9K_{s}.$ So, if $K_{s} = K_{1} = K$, then
$K_{Cu} = K_{2} = 9K$
$\Rightarrow$ $\theta = \frac{9K \times 100 \times 6 + K \times 0 \times 18}{9K \times 6 + K \times 18}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{5400K}{72K} = 75℃$